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The crude monadicity theorem in its reflexive-coequalizer form says: if $F⊣G:\mathcal{D}\to \mathcal{C}$, $G$ is a conservative functor, $\mathcal{D}$ has reflexive coequalizers, and $G$ preserves them, then the Eilenberg-Moore comparison functor $K:\mathcal{D}\to {\mathcal{C}}^T$, for $T=GF$, is an equivalence of categories. Requiring all coequalizers to exist and be preserved is a stronger sufficient form.

For an algebra $(A,a)$, form in $\mathcal{D}$ the coequalizerThe pair is a reflexive pair with common section $F{\eta }_A$: both composites are the identity by the algebra unit law and the triangle identity. An arrow $u:FA\to B$ transposes to $v=Gu{\eta }_A:A\to GB$. The two composites $uFa$ and $u{\epsilon }_{FA}$ transpose to $va$ and $G{\epsilon }_{B}Tv$, respectively. Indeed the latter transpose is $Gu{\mu }_A{\eta }_{TA}=Gu$, while $G{\epsilon }_{B}Tv=Gu{\mu }_{A}T{\eta }_A=Gu$ by counit naturality. Thus $u$ equalizes the pair exactly when $v$ is an monad algebra morphism $(A,a)\to K(B)$. The coequalizer property gives natural bijectionsThese define $L$ on arrows by uniqueness, so $L⊣K$. Its unit has underlying arrow $\beta =Gq{\eta }_A:A\to GL(A,a)$.

By preservation, $Gq$ coequalizes $Ta$ and ${\mu }_A$ in $\mathcal{C}$. The action $a:TA\to A$ is a split coequalizer of that pair: take $s={\eta }_A$ and $t={\eta }_{TA}$, with ${\mu }_{A}t=1_{TA}$ and $Tat={\eta }_{A}a$. Thus there is a unique isomorphism $\alpha :GL(A,a)\to A$ with $\alpha Gq=a$. We have $\alpha \beta =a{\eta }_A=1_A$. AlsoConsequently $\beta \alpha Gq=Gq$, and epimorphic cancellation gives $\beta \alpha =1$. The adjunction unit $\beta$ is an monad algebra morphism; its invertible underlying arrow has an algebra-morphism inverse. Thus the unit of $L⊣K$ is an isomorphism.

For its counit $c_B:LK(B)\to B$, the triangle identity gives $K(c_B){\beta }_{K(B)}=1_{K(B)}$. Hence $K(c_B)$, and therefore $G(c_B)$, is an isomorphism. Since $G$ is a conservative functor, $c_B$ is an isomorphism too. Both unit and counit of $L⊣K$ are invertible, which proves the claimed equivalence. This proof uses only reflexive coequalizers in $\mathcal{D}$ and split coequalizers in $\mathcal{C}$.

Take the definition of an elementary topos as a category with finite limits, exponential objects and a subobject classifier $\top :1\hookrightarrow \Omega$. Its power object is $PX={\Omega }^X$, and $Pf:PY\to PX$ is precomposition with $f:X\to Y$. Transposing a predicate on $X\times Y$ in either variable givesThus $P^{\mathrm{op}}:\mathcal{E}\to {\mathcal{E}}^{\mathrm{op}}$ is left adjoint to the contravariant power-object functor $P:{\mathcal{E}}^{\mathrm{op}}\to \mathcal{E}$.

Here are the remaining hypotheses for the crude monadicity theorem, obtained from those topos axioms. The unit of this adjunction is ${\eta }_X:X\to PPX$, internally ${\eta }_X(x)(S)=(x\in S)$. It is monic: if two such evaluations agree, evaluate on the singleton predicate $\{x\}$ classified by the diagonal to obtain $x=y$. If $Pf$ is invertible, naturality $PPf{\eta }_X={\eta }_{Y}f$ therefore makes $f$ monic. The characteristic map of this mono, regarded as a global element of $PY$, pulls back along $f$ to the everywhere-true predicate on $X$. Since $Pf$ is monic, that characteristic map was already everywhere true on $Y$. The classifier pullback then says that $f$ is invertible. Hence $P$ is a conservative functor.

It remains to check preservation of reflexive coequalizers in the opposite category. Equivalently, let $f,g:B\rightrightarrows A$ be a coreflexive pair, with $r:A\to B$ satisfying $rf=rg=1_B$, and let $e:E\hookrightarrow B$ be their equalizer. Both $f$ and $g$ are monic. For any mono $i$, there is a direct-image map ${\exists }_i$ between power objects: a subobject is sent to its composite with $i$. This uses only classification of monos, not the prior existence of general images or colimits. We have $Pi{\exists }_i=1$.

For a predicate $W\hookrightarrow B$, consider its direct image along $f$. Its pullbacks along the two sections areTo verify the second equality, $g(b)=f(w)$ with $w\in W$ implies $b=w$ after applying $r$, and therefore $f(b)=g(b)$; conversely $b\in W\cap E$ supplies that witness. This argument works for parameterized subobjects as well, so it is an equality of arrows between power objects.

Now if $h:PB\to Z$ satisfies $hPf=hPg$, then $h=h{\exists }_{e}Pe$. Hence $h{\exists }_e:PE\to Z$ is its unique factorization through $Pe$, uniqueness following from the section ${\exists }_e$. Thereforeis a coequalizer. Finite limits supply every required coreflexive equalizer. The right adjoint $P$ reflects isomorphisms and preserves the corresponding reflexive coequalizers, so $P$ is monadic. In particular ${\mathcal{E}}^{\mathrm{op}}$ is equivalent to the Eilenberg-Moore category of the double-power-object monad on $\mathcal{E}$.

Let $F:\mathcal{E}\to \mathcal{F}$ be a logical functor, and suppose $L⊣F$. Preservation of exponentials and the classifier gives $F{P}_{\mathcal{E}}\cong P_{\mathcal{F}}{F}^{\mathrm{op}}$, compatibly with the units, counits and resulting monads. Thus $F^{\mathrm{op}}$ is the algebra functor lifting the base functor $F$ through the two monadic power-object functors. The adjoint lifting theorem for monad algebra functors applies to the base adjunction $L⊣F$: its required coequalizers exist in ${\mathcal{E}}^{\mathrm{op}}$, because $\mathcal{E}$ has finite equalizers. It supplies a left adjoint $K⊣F^{\mathrm{op}}$. Taking opposites gives $F⊣K^{\mathrm{op}}$, the required right adjoint to $F$.

Preserving either of the two logical structures by itself is insufficient. For the exponential example take $F:\mathbf{Set}\to \mathbf{Set}$ constant at $1$. The constant-empty functor is its left adjoint, since both relevant hom-sets are singletons. Its canonical exponential comparison is $1\cong 1^1$, so it preserves exponentials. It has no right adjoint: a functor with a right adjoint would preserve the initial object, whereas $F(\varnothing )=1$.

For the classifier example take $G:(\mathbb{Z}/2)\text{-}\mathbf{Set}\to \mathbf{Set}$ to be fixed points. The trivial-action functor is left adjoint to $G$. In a group-action topos the classifier is the trivial-action two-element set, since invariant subsets have ordinary equivariant characteristic maps. Therefore $G$ preserves the classifier, its true arrow and the terminal object. But $G$ does not preserve the coequalizer of the identity and the nontrivial translation on the regular two-element group set: that coequalizer is $1$, while the fixed-point sets of the domain and codomain of the parallel pair are empty. Their set-theoretic coequalizer is empty, not $G(1)=1$. Thus $G$ has no right adjoint.